Let the consecutive numbers be $x$, $x+1$, and $x+2$. There are three cases based on which of the numbers is the sum of the other two.

If $x+2 = x + (x+1)$, then $x = 1$, which gives $(1, 2, 3)$.

If $x+1 = x + (x+2)$, then $x = -1$, which gives $(-1, 0, 1)$.

If $x = (x+1) + (x+2)$, then $x = -3$, which gives $(-3, -2, -1)$.

If all three numbers are positive, then only the greatest of them can be the sum of the two others. As the sum has to be greater than one of the summands by 1, the other summand has to be 1. This results in the triplet $(1, 2, 3)$.

If all the three numbers are negative, then only the smallest of the three numbers can be the sum of the other two. As the sum has to be smaller than one of the summands by 1, the other summand has to be $-1$. This results in the triplet $(-3, -2, -1)$.

If the numbers are neither all positive nor all negative, then they have to include $0$. Number $0$ cannot be a summand, otherwise the sum would equal the other summand. Hence $0$ has to be the sum of the other numbers. In that case the summands must be $-1$ and $1$. This results in the final triplet $(-1, 0, 1)$.

One of the summands has to be either greater than the sum by $1$ or less than the sum by $1$. Hence the second summand is equal to the difference of the sum and the first summand, so it is either $1$ or $-1$. It cannot be the middle number in the triplet (otherwise the difference of the other two numbers cannot be $1$ or $-1$). Hence, we only need to investigate the triplets in which either the smallest or the largest number is either $1$ or $-1$. The only such triplets are $(1, 2, 3)$, $(-1, 0, 1)$, and $(-3, -2, -1)$. All of them result in solutions, as $1+2=3$, $-1+1=0$, and $-2+(-1)=-3$.